3.1.15 \(\int x^3 (a+b \sec ^{-1}(c x))^2 \, dx\) [15]

Optimal. Leaf size=107 \[ \frac {b^2 x^2}{12 c^2}-\frac {b \sqrt {1-\frac {1}{c^2 x^2}} x \left (a+b \sec ^{-1}(c x)\right )}{3 c^3}-\frac {b \sqrt {1-\frac {1}{c^2 x^2}} x^3 \left (a+b \sec ^{-1}(c x)\right )}{6 c}+\frac {1}{4} x^4 \left (a+b \sec ^{-1}(c x)\right )^2+\frac {b^2 \log (x)}{3 c^4} \]

[Out]

1/12*b^2*x^2/c^2+1/4*x^4*(a+b*arcsec(c*x))^2+1/3*b^2*ln(x)/c^4-1/3*b*x*(a+b*arcsec(c*x))*(1-1/c^2/x^2)^(1/2)/c
^3-1/6*b*x^3*(a+b*arcsec(c*x))*(1-1/c^2/x^2)^(1/2)/c

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Rubi [A]
time = 0.08, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {5330, 4494, 4270, 4269, 3556} \begin {gather*} -\frac {b x^3 \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )}{6 c}-\frac {b x \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )}{3 c^3}+\frac {1}{4} x^4 \left (a+b \sec ^{-1}(c x)\right )^2+\frac {b^2 \log (x)}{3 c^4}+\frac {b^2 x^2}{12 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*ArcSec[c*x])^2,x]

[Out]

(b^2*x^2)/(12*c^2) - (b*Sqrt[1 - 1/(c^2*x^2)]*x*(a + b*ArcSec[c*x]))/(3*c^3) - (b*Sqrt[1 - 1/(c^2*x^2)]*x^3*(a
+ b*ArcSec[c*x]))/(6*c) + (x^4*(a + b*ArcSec[c*x])^2)/4 + (b^2*Log[x])/(3*c^4)

Rule 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 4269

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(c + d*x)^m)*(Cot[e + f*x]/f), x
] + Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 4270

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(-b^2)*(c + d*x)*Cot[e + f*x]
*((b*Csc[e + f*x])^(n - 2)/(f*(n - 1))), x] + (Dist[b^2*((n - 2)/(n - 1)), Int[(c + d*x)*(b*Csc[e + f*x])^(n -
2), x], x] - Simp[b^2*d*((b*Csc[e + f*x])^(n - 2)/(f^2*(n - 1)*(n - 2))), x]) /; FreeQ[{b, c, d, e, f}, x] &&
GtQ[n, 1] && NeQ[n, 2]

Rule 4494

Int[((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.)*Tan[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Simp[
(c + d*x)^m*(Sec[a + b*x]^n/(b*n)), x] - Dist[d*(m/(b*n)), Int[(c + d*x)^(m - 1)*Sec[a + b*x]^n, x], x] /; Fre
eQ[{a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[m, 0]

Rule 5330

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[(a + b*x)^n*S
ec[x]^(m + 1)*Tan[x], x], x, ArcSec[c*x]], x] /; FreeQ[{a, b, c}, x] && IntegerQ[n] && IntegerQ[m] && (GtQ[n,
0] || LtQ[m, -1])

Rubi steps

\begin {align*} \int x^3 \left (a+b \sec ^{-1}(c x)\right )^2 \, dx &=\frac {\text {Subst}\left (\int (a+b x)^2 \sec ^4(x) \tan (x) \, dx,x,\sec ^{-1}(c x)\right )}{c^4}\\ &=\frac {1}{4} x^4 \left (a+b \sec ^{-1}(c x)\right )^2-\frac {b \text {Subst}\left (\int (a+b x) \sec ^4(x) \, dx,x,\sec ^{-1}(c x)\right )}{2 c^4}\\ &=\frac {b^2 x^2}{12 c^2}-\frac {b \sqrt {1-\frac {1}{c^2 x^2}} x^3 \left (a+b \sec ^{-1}(c x)\right )}{6 c}+\frac {1}{4} x^4 \left (a+b \sec ^{-1}(c x)\right )^2-\frac {b \text {Subst}\left (\int (a+b x) \sec ^2(x) \, dx,x,\sec ^{-1}(c x)\right )}{3 c^4}\\ &=\frac {b^2 x^2}{12 c^2}-\frac {b \sqrt {1-\frac {1}{c^2 x^2}} x \left (a+b \sec ^{-1}(c x)\right )}{3 c^3}-\frac {b \sqrt {1-\frac {1}{c^2 x^2}} x^3 \left (a+b \sec ^{-1}(c x)\right )}{6 c}+\frac {1}{4} x^4 \left (a+b \sec ^{-1}(c x)\right )^2+\frac {b^2 \text {Subst}\left (\int \tan (x) \, dx,x,\sec ^{-1}(c x)\right )}{3 c^4}\\ &=\frac {b^2 x^2}{12 c^2}-\frac {b \sqrt {1-\frac {1}{c^2 x^2}} x \left (a+b \sec ^{-1}(c x)\right )}{3 c^3}-\frac {b \sqrt {1-\frac {1}{c^2 x^2}} x^3 \left (a+b \sec ^{-1}(c x)\right )}{6 c}+\frac {1}{4} x^4 \left (a+b \sec ^{-1}(c x)\right )^2+\frac {b^2 \log (x)}{3 c^4}\\ \end {align*}

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Mathematica [A]
time = 0.14, size = 124, normalized size = 1.16 \begin {gather*} \frac {c x \left (b^2 c x+3 a^2 c^3 x^3-2 a b \sqrt {1-\frac {1}{c^2 x^2}} \left (2+c^2 x^2\right )\right )-2 b c x \left (-3 a c^3 x^3+b \sqrt {1-\frac {1}{c^2 x^2}} \left (2+c^2 x^2\right )\right ) \sec ^{-1}(c x)+3 b^2 c^4 x^4 \sec ^{-1}(c x)^2+4 b^2 \log (x)}{12 c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*ArcSec[c*x])^2,x]

[Out]

(c*x*(b^2*c*x + 3*a^2*c^3*x^3 - 2*a*b*Sqrt[1 - 1/(c^2*x^2)]*(2 + c^2*x^2)) - 2*b*c*x*(-3*a*c^3*x^3 + b*Sqrt[1
- 1/(c^2*x^2)]*(2 + c^2*x^2))*ArcSec[c*x] + 3*b^2*c^4*x^4*ArcSec[c*x]^2 + 4*b^2*Log[x])/(12*c^4)

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Maple [A]
time = 0.25, size = 181, normalized size = 1.69

method	result	size
derivativedivides	\(\frac {\frac {c^{4} x^{4} a^{2}}{4}+\frac {b^{2} \mathrm {arcsec}\left (c x \right )^{2} c^{4} x^{4}}{4}-\frac {b^{2} \mathrm {arcsec}\left (c x \right ) \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c^{3} x^{3}}{6}+\frac {b^{2} c^{2} x^{2}}{12}-\frac {b^{2} \mathrm {arcsec}\left (c x \right ) c x \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}}{3}-\frac {b^{2} \ln \left (\frac {1}{c x}\right )}{3}+2 a b \left (\frac {c^{4} x^{4} \mathrm {arcsec}\left (c x \right )}{4}-\frac {\left (c^{2} x^{2}-1\right ) \left (c^{2} x^{2}+2\right )}{12 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c x}\right )}{c^{4}}\)	\(181\)
default	\(\frac {\frac {c^{4} x^{4} a^{2}}{4}+\frac {b^{2} \mathrm {arcsec}\left (c x \right )^{2} c^{4} x^{4}}{4}-\frac {b^{2} \mathrm {arcsec}\left (c x \right ) \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c^{3} x^{3}}{6}+\frac {b^{2} c^{2} x^{2}}{12}-\frac {b^{2} \mathrm {arcsec}\left (c x \right ) c x \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}}{3}-\frac {b^{2} \ln \left (\frac {1}{c x}\right )}{3}+2 a b \left (\frac {c^{4} x^{4} \mathrm {arcsec}\left (c x \right )}{4}-\frac {\left (c^{2} x^{2}-1\right ) \left (c^{2} x^{2}+2\right )}{12 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c x}\right )}{c^{4}}\)	\(181\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arcsec(c*x))^2,x,method=_RETURNVERBOSE)

[Out]

1/c^4*(1/4*c^4*x^4*a^2+1/4*b^2*arcsec(c*x)^2*c^4*x^4-1/6*b^2*arcsec(c*x)*((c^2*x^2-1)/c^2/x^2)^(1/2)*c^3*x^3+1
/12*b^2*c^2*x^2-1/3*b^2*arcsec(c*x)*c*x*((c^2*x^2-1)/c^2/x^2)^(1/2)-1/3*b^2*ln(1/c/x)+2*a*b*(1/4*c^4*x^4*arcse
c(c*x)-1/12*(c^2*x^2-1)*(c^2*x^2+2)/((c^2*x^2-1)/c^2/x^2)^(1/2)/c/x))

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Maxima [A]
time = 0.52, size = 163, normalized size = 1.52 \begin {gather*} \frac {1}{4} \, b^{2} x^{4} \operatorname {arcsec}\left (c x\right )^{2} + \frac {1}{4} \, a^{2} x^{4} + \frac {1}{6} \, {\left (3 \, x^{4} \operatorname {arcsec}\left (c x\right ) - \frac {c^{2} x^{3} {\left (-\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {3}{2}} + 3 \, x \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{c^{3}}\right )} a b + \frac {{\left ({\left (c^{2} x^{2} + 2 \, \log \left (x^{2}\right )\right )} \sqrt {c x + 1} \sqrt {c x - 1} - 2 \, {\left (c^{4} x^{4} + c^{2} x^{2} - 2\right )} \arctan \left (\sqrt {c x + 1} \sqrt {c x - 1}\right )\right )} b^{2}}{12 \, \sqrt {c x + 1} \sqrt {c x - 1} c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arcsec(c*x))^2,x, algorithm="maxima")

[Out]

1/4*b^2*x^4*arcsec(c*x)^2 + 1/4*a^2*x^4 + 1/6*(3*x^4*arcsec(c*x) - (c^2*x^3*(-1/(c^2*x^2) + 1)^(3/2) + 3*x*sqr
t(-1/(c^2*x^2) + 1))/c^3)*a*b + 1/12*((c^2*x^2 + 2*log(x^2))*sqrt(c*x + 1)*sqrt(c*x - 1) - 2*(c^4*x^4 + c^2*x^
2 - 2)*arctan(sqrt(c*x + 1)*sqrt(c*x - 1)))*b^2/(sqrt(c*x + 1)*sqrt(c*x - 1)*c^4)

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Fricas [A]
time = 5.18, size = 146, normalized size = 1.36 \begin {gather*} \frac {3 \, b^{2} c^{4} x^{4} \operatorname {arcsec}\left (c x\right )^{2} + 3 \, a^{2} c^{4} x^{4} + 12 \, a b c^{4} \arctan \left (-c x + \sqrt {c^{2} x^{2} - 1}\right ) + b^{2} c^{2} x^{2} + 4 \, b^{2} \log \left (x\right ) + 6 \, {\left (a b c^{4} x^{4} - a b c^{4}\right )} \operatorname {arcsec}\left (c x\right ) - 2 \, {\left (a b c^{2} x^{2} + 2 \, a b + {\left (b^{2} c^{2} x^{2} + 2 \, b^{2}\right )} \operatorname {arcsec}\left (c x\right )\right )} \sqrt {c^{2} x^{2} - 1}}{12 \, c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arcsec(c*x))^2,x, algorithm="fricas")

[Out]

1/12*(3*b^2*c^4*x^4*arcsec(c*x)^2 + 3*a^2*c^4*x^4 + 12*a*b*c^4*arctan(-c*x + sqrt(c^2*x^2 - 1)) + b^2*c^2*x^2
+ 4*b^2*log(x) + 6*(a*b*c^4*x^4 - a*b*c^4)*arcsec(c*x) - 2*(a*b*c^2*x^2 + 2*a*b + (b^2*c^2*x^2 + 2*b^2)*arcsec
(c*x))*sqrt(c^2*x^2 - 1))/c^4

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{3} \left (a + b \operatorname {asec}{\left (c x \right )}\right )^{2}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*asec(c*x))**2,x)

[Out]

Integral(x**3*(a + b*asec(c*x))**2, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 6625 vs. \(2 (93) = 186\).
time = 0.73, size = 6625, normalized size = 61.92 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arcsec(c*x))^2,x, algorithm="giac")

[Out]

1/12*(3*b^2*arccos(1/(c*x))^2/(c^5 + 4*c^5*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 6*c^5*(1/(c^2*x^2) - 1)^2/(1/(c
*x) + 1)^4 + 4*c^5*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + c^5*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8) + 6*a*b*arcc
os(1/(c*x))/(c^5 + 4*c^5*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 6*c^5*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 4*c^5
*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + c^5*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8) - 12*b^2*(1/(c^2*x^2) - 1)*arc
cos(1/(c*x))^2/((c^5 + 4*c^5*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 6*c^5*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 4
*c^5*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + c^5*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8)*(1/(c*x) + 1)^2) - 4*b^2*l
og(2)/(c^5 + 4*c^5*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 6*c^5*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 4*c^5*(1/(c
^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + c^5*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8) + 4*b^2*log(2/(c*x) + 2)/(c^5 + 4*c^
5*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 6*c^5*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 4*c^5*(1/(c^2*x^2) - 1)^3/(1
/(c*x) + 1)^6 + c^5*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8) - 4*b^2*log(abs(sqrt(-1/(c^2*x^2) + 1) + 1/(c*x) + 1)
)/(c^5 + 4*c^5*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 6*c^5*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 4*c^5*(1/(c^2*x
^2) - 1)^3/(1/(c*x) + 1)^6 + c^5*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8) - 4*b^2*log(abs(sqrt(-1/(c^2*x^2) + 1) -
1/(c*x) - 1))/(c^5 + 4*c^5*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 6*c^5*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 4*
c^5*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + c^5*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8) - 12*b^2*sqrt(-1/(c^2*x^2)
+ 1)*arccos(1/(c*x))/((c^5 + 4*c^5*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 6*c^5*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)
^4 + 4*c^5*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + c^5*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8)*(1/(c*x) + 1)) + 3*a
^2/(c^5 + 4*c^5*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 6*c^5*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 4*c^5*(1/(c^2*
x^2) - 1)^3/(1/(c*x) + 1)^6 + c^5*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8) + b^2/(c^5 + 4*c^5*(1/(c^2*x^2) - 1)/(1
/(c*x) + 1)^2 + 6*c^5*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 4*c^5*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + c^5*(1
/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8) - 24*a*b*(1/(c^2*x^2) - 1)*arccos(1/(c*x))/((c^5 + 4*c^5*(1/(c^2*x^2) - 1)/
(1/(c*x) + 1)^2 + 6*c^5*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 4*c^5*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + c^5*
(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8)*(1/(c*x) + 1)^2) + 18*b^2*(1/(c^2*x^2) - 1)^2*arccos(1/(c*x))^2/((c^5 + 4
*c^5*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 6*c^5*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 4*c^5*(1/(c^2*x^2) - 1)^3
/(1/(c*x) + 1)^6 + c^5*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8)*(1/(c*x) + 1)^4) - 16*b^2*(1/(c^2*x^2) - 1)*log(2)
/((c^5 + 4*c^5*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 6*c^5*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 4*c^5*(1/(c^2*x
^2) - 1)^3/(1/(c*x) + 1)^6 + c^5*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8)*(1/(c*x) + 1)^2) + 16*b^2*(1/(c^2*x^2) -
1)*log(2/(c*x) + 2)/((c^5 + 4*c^5*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 6*c^5*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)
^4 + 4*c^5*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + c^5*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8)*(1/(c*x) + 1)^2) - 1
6*b^2*(1/(c^2*x^2) - 1)*log(abs(sqrt(-1/(c^2*x^2) + 1) + 1/(c*x) + 1))/((c^5 + 4*c^5*(1/(c^2*x^2) - 1)/(1/(c*x
) + 1)^2 + 6*c^5*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 4*c^5*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + c^5*(1/(c^2
*x^2) - 1)^4/(1/(c*x) + 1)^8)*(1/(c*x) + 1)^2) - 16*b^2*(1/(c^2*x^2) - 1)*log(abs(sqrt(-1/(c^2*x^2) + 1) - 1/(
c*x) - 1))/((c^5 + 4*c^5*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 6*c^5*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 4*c^5
*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + c^5*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8)*(1/(c*x) + 1)^2) - 12*a*b*sqrt
(-1/(c^2*x^2) + 1)/((c^5 + 4*c^5*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 6*c^5*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4
+ 4*c^5*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + c^5*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8)*(1/(c*x) + 1)) + 20*b^
2*(-1/(c^2*x^2) + 1)^(3/2)*arccos(1/(c*x))/((c^5 + 4*c^5*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 6*c^5*(1/(c^2*x^2
) - 1)^2/(1/(c*x) + 1)^4 + 4*c^5*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + c^5*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8
)*(1/(c*x) + 1)^3) - 12*a^2*(1/(c^2*x^2) - 1)/((c^5 + 4*c^5*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 6*c^5*(1/(c^2*
x^2) - 1)^2/(1/(c*x) + 1)^4 + 4*c^5*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + c^5*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1
)^8)*(1/(c*x) + 1)^2) + 36*a*b*(1/(c^2*x^2) - 1)^2*arccos(1/(c*x))/((c^5 + 4*c^5*(1/(c^2*x^2) - 1)/(1/(c*x) +
1)^2 + 6*c^5*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 4*c^5*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + c^5*(1/(c^2*x^2
) - 1)^4/(1/(c*x) + 1)^8)*(1/(c*x) + 1)^4) - 12*b^2*(1/(c^2*x^2) - 1)^3*arccos(1/(c*x))^2/((c^5 + 4*c^5*(1/(c^
2*x^2) - 1)/(1/(c*x) + 1)^2 + 6*c^5*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 4*c^5*(1/(c^2*x^2) - 1)^3/(1/(c*x) +
1)^6 + c^5*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8)*(1/(c*x) + 1)^6) - 24*b^2*(1/(c^2*x^2) - 1)^2*log(2)/((c^5 +
4*c^5*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 6*c^5*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 4*c^5*(1/(c^2*x^2) - 1)^
3/(1/(c*x) + 1)^6 + c^5*(1/(c^2*x^2) - 1)^4/(1/...

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^3\,{\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*acos(1/(c*x)))^2,x)

[Out]

int(x^3*(a + b*acos(1/(c*x)))^2, x)

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